\(\int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx\) [38]

Optimal result

Integrand size = 23, antiderivative size = 251 \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx=\frac {(c+d x)^{1+m}}{8 a^3 d (1+m)}+\frac {3 i 2^{-4-m} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{a^3 f}+\frac {3 i 2^{-5-2 m} e^{-4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i f (c+d x)}{d}\right )}{a^3 f}+\frac {i 2^{-4-m} 3^{-1-m} e^{-6 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {6 i f (c+d x)}{d}\right )}{a^3 f} \] Output:

1/8*(d*x+c)^(1+m)/a^3/d/(1+m)+3*I*2^(-4-m)*(d*x+c)^m*GAMMA(1+m,2*I*f*(d*x+ 
c)/d)/a^3/exp(2*I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m)+3*I*2^(-5-2*m)*(d*x+c)^ 
m*GAMMA(1+m,4*I*f*(d*x+c)/d)/a^3/exp(4*I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m)+ 
I*2^(-4-m)*3^(-1-m)*(d*x+c)^m*GAMMA(1+m,6*I*f*(d*x+c)/d)/a^3/exp(6*I*(e-c* 
f/d))/f/((I*f*(d*x+c)/d)^m)
 

Mathematica [A] (verified)

Time = 12.42 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx=\frac {2^{-5-2 m} 3^{-1-m} e^{-3 i e} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \left (12^{1+m} e^{6 i e} f (c+d x) \left (\frac {i f (c+d x)}{d}\right )^m+i 2^{1+m} 3^{2+m} d e^{2 i \left (2 e+\frac {c f}{d}\right )} (1+m) \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )+i 3^{2+m} d e^{2 i e+\frac {4 i c f}{d}} (1+m) \Gamma \left (1+m,\frac {4 i f (c+d x)}{d}\right )+i 2^{1+m} d e^{\frac {6 i c f}{d}} (1+m) \Gamma \left (1+m,\frac {6 i f (c+d x)}{d}\right )\right ) \sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3}{d f (1+m) (a+i a \tan (e+f x))^3} \] Input:

Integrate[(c + d*x)^m/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

(2^(-5 - 2*m)*3^(-1 - m)*(c + d*x)^m*(12^(1 + m)*E^((6*I)*e)*f*(c + d*x)*( 
(I*f*(c + d*x))/d)^m + I*2^(1 + m)*3^(2 + m)*d*E^((2*I)*(2*e + (c*f)/d))*( 
1 + m)*Gamma[1 + m, ((2*I)*f*(c + d*x))/d] + I*3^(2 + m)*d*E^((2*I)*e + (( 
4*I)*c*f)/d)*(1 + m)*Gamma[1 + m, ((4*I)*f*(c + d*x))/d] + I*2^(1 + m)*d*E 
^(((6*I)*c*f)/d)*(1 + m)*Gamma[1 + m, ((6*I)*f*(c + d*x))/d])*Sec[e + f*x] 
^3*(Cos[f*x] + I*Sin[f*x])^3)/(d*E^((3*I)*e)*f*(1 + m)*((I*f*(c + d*x))/d) 
^m*(a + I*a*Tan[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4212, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4212

\(\displaystyle \int \left (\frac {3 e^{-2 i e-2 i f x} (c+d x)^m}{8 a^3}+\frac {3 e^{-4 i e-4 i f x} (c+d x)^m}{8 a^3}+\frac {e^{-6 i e-6 i f x} (c+d x)^m}{8 a^3}+\frac {(c+d x)^m}{8 a^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 i 2^{-m-4} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{a^3 f}+\frac {3 i 2^{-2 m-5} e^{-4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 i f (c+d x)}{d}\right )}{a^3 f}+\frac {i 2^{-m-4} 3^{-m-1} e^{-6 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {6 i f (c+d x)}{d}\right )}{a^3 f}+\frac {(c+d x)^{m+1}}{8 a^3 d (m+1)}\)

Input:

Int[(c + d*x)^m/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

(c + d*x)^(1 + m)/(8*a^3*d*(1 + m)) + ((3*I)*2^(-4 - m)*(c + d*x)^m*Gamma[ 
1 + m, ((2*I)*f*(c + d*x))/d])/(a^3*E^((2*I)*(e - (c*f)/d))*f*((I*f*(c + d 
*x))/d)^m) + ((3*I)*2^(-5 - 2*m)*(c + d*x)^m*Gamma[1 + m, ((4*I)*f*(c + d* 
x))/d])/(a^3*E^((4*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) + (I*2^(-4 - 
 m)*3^(-1 - m)*(c + d*x)^m*Gamma[1 + m, ((6*I)*f*(c + d*x))/d])/(a^3*E^((6 
*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4212

Int[((c_.) + (d_.)*(x_))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), 
x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (1/(2*a) + E^(2*(a/b)*(e + f* 
x))/(2*a))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 + b^2 
, 0] && ILtQ[n, 0]
 

Maple [F]

\[\int \frac {\left (d x +c \right )^{m}}{\left (a +i a \tan \left (f x +e \right )\right )^{3}}d x\]

Input:

int((d*x+c)^m/(a+I*a*tan(f*x+e))^3,x)
 

Output:

int((d*x+c)^m/(a+I*a*tan(f*x+e))^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.79 \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx=-\frac {18 \, {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) + 9 \, {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {4 i \, f}{d}\right ) + 4 i \, d e - 4 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) + 2 \, {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {6 i \, f}{d}\right ) + 6 i \, d e - 6 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {6 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - 12 \, {\left (d f x + c f\right )} {\left (d x + c\right )}^{m}}{96 \, {\left (a^{3} d f m + a^{3} d f\right )}} \] Input:

integrate((d*x+c)^m/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

-1/96*(18*(-I*d*m - I*d)*e^(-(d*m*log(2*I*f/d) + 2*I*d*e - 2*I*c*f)/d)*gam 
ma(m + 1, -2*(-I*d*f*x - I*c*f)/d) + 9*(-I*d*m - I*d)*e^(-(d*m*log(4*I*f/d 
) + 4*I*d*e - 4*I*c*f)/d)*gamma(m + 1, -4*(-I*d*f*x - I*c*f)/d) + 2*(-I*d* 
m - I*d)*e^(-(d*m*log(6*I*f/d) + 6*I*d*e - 6*I*c*f)/d)*gamma(m + 1, -6*(-I 
*d*f*x - I*c*f)/d) - 12*(d*f*x + c*f)*(d*x + c)^m)/(a^3*d*f*m + a^3*d*f)
 

Sympy [F]

\[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {\left (c + d x\right )^{m}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \] Input:

integrate((d*x+c)**m/(a+I*a*tan(f*x+e))**3,x)
 

Output:

I*Integral((c + d*x)**m/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + 
 f*x) + I), x)/a**3
 

Maxima [F]

\[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx=\int { \frac {{\left (d x + c\right )}^{m}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((d*x+c)^m/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

1/8*((d*m + d)*integrate((d*x + c)^m*cos(6*f*x + 6*e), x) + 3*(d*m + d)*in 
tegrate((d*x + c)^m*cos(4*f*x + 4*e), x) + 3*(d*m + d)*integrate((d*x + c) 
^m*cos(2*f*x + 2*e), x) - (I*d*m + I*d)*integrate((d*x + c)^m*sin(6*f*x + 
6*e), x) + 3*(-I*d*m - I*d)*integrate((d*x + c)^m*sin(4*f*x + 4*e), x) + 3 
*(-I*d*m - I*d)*integrate((d*x + c)^m*sin(2*f*x + 2*e), x) + e^(m*log(d*x 
+ c) + log(d*x + c)))/(a^3*d*m + a^3*d)
 

Giac [F]

\[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx=\int { \frac {{\left (d x + c\right )}^{m}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((d*x+c)^m/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

integrate((d*x + c)^m/(I*a*tan(f*x + e) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx=\int \frac {{\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \] Input:

int((c + d*x)^m/(a + a*tan(e + f*x)*1i)^3,x)
 

Output:

int((c + d*x)^m/(a + a*tan(e + f*x)*1i)^3, x)
 

Reduce [F]

\[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx=-\frac {\int \frac {\left (d x +c \right )^{m}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x}{a^{3}} \] Input:

int((d*x+c)^m/(a+I*a*tan(f*x+e))^3,x)
 

Output:

( - int((c + d*x)**m/(tan(e + f*x)**3*i + 3*tan(e + f*x)**2 - 3*tan(e + f* 
x)*i - 1),x))/a**3